The X-axis is Log10 and Y-axis is dB. Why does the cutoff frequency differ between the simulation and on paper? Am I calculating it wrong or is there something wrong with the simulation code? How can I calculate the lower cutoff frequency" for this circuit? Also, how can I calculate the low frequency poles for this circuit? Your method of going from an open-loop bandwidth to a closed-loop bandwidth is wrong. If the unity gain open-loop bandwidth is 1 MHz then for a closed-loop gain of 10 the closed-loop bandwidth will be kHz.
In more detail: as the frequency approaches kHz the actual gain starts to drop from 10 to 7. Picture from here. Because there is a capacitor C1 present, does it mean the circuit should have a lower cutoff frequency? Yes, certainly - you now have two stages of filtering and they are multiplicative in that their respective transfer functions can be multiplied together to get the overall transfer function.
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Asked 2 years, 1 month ago. Modified 2 years, 1 month ago. Viewed times. Step 2 - Frequency Response I will now try to calculate the closed-loop gain of the circuit. This resistor is used to balance out asymmetries in real op-amps caused by transistor mismatch. Since simulations ignore the random variations between devices that is introduced in manufacturing, there is no need for this resistor in simulations. Just leave it out.
Questions Why does the cutoff frequency differ between the simulation and on paper? Ali Nategh. Ali Nategh Ali Nategh 78 8 8 bronze badges. Show 4 more comments. Sorted by: Reset to default. Highest score default Date modified newest first Date created oldest first. For a closed-loop gain of 11, the 3 dB point in the response will be around 91 kHz. Andy aka Andy aka k 23 23 gold badges silver badges bronze badges. With amplifiers configured for a high closed loop gain, one will primarily see the effect of the limited over all bandwidth.
With low gain amplifiers the details of the feedback circuit become important, as well as the behavior of the op-amp at high frequency. Stray capacitance can cause phase shifts that make the circuit unstable. Apply a square wave, kHz, mV signal to the input and compare input and output signals.
Do this for the LM and the LT The oscillations observed with the LT can be eliminated with a small capacitor parallel to the feedback resistor. Figure 3. A circuit to determine the transient response. Secondly, increase the gain of the amplifier to 10, and again compare the behavior of the circuit with an LM and an LT It should be clear that while the LT can follow the input signal with ease, the LM is now too slow.
Finally, still with an amplifier gain of ten, look at what happens when the signal amplitude is increased. For small amplitude, you should observe a well defined, amplitude independent, rise time for the LM When the input voltage exceeds mV the character of the output voltage starts to change. It now appears that the voltage can only change at a fixed rate, independent of the amplitude at the input.
This rate is called the slew rate. Try to repeat this with the LT Non-ideal op-amps Download to PDF 0. Introduction In our analysis of circuits based on the operational amplifier, we made a number of important assumptions: a. Input offset voltage and noise Assemble a non-inverting amplifier with a gain of , Fig. Figure 1. Offset and noise 2. Input currents To measure the input current you can connect a resistor, R in , between the non-inverting input and ground. Determining the bias current for the LT 3.
Gain-bandwidth product A convenient number to characterize an op-amp is its gain-bandwidth product. Transient response and slew rate The transient response, i.
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For the op-amp. The high gain and wide range of operating voltage provide superior performance in integrator, summing amplifier, and general feedback applications. The LM A type OPAMP has a gain bandwidth product of 1 MHz. A non-inverting amplifier using this OPAMP and having a voltage gain of 20 dB will exhibit a dB.